Find the minimum value of (5 cosh x + 3 sinh x).
How can I solve this?
Differentiate, put to zero, test nature .... the usual deal.
Note that to solve when the derivative is equal to zero, you should:
1. Replace cosh and sinh with their exponential definitions.
2. Multiply through by 2 e^{-x}.
3. Solve the resulting quadratic equation in e^x .....
$\displaystyle \cosh x = \frac{e^x + e^{-x}}{2}$
$\displaystyle \sinh x = \frac{e^x - e^{-x}}{2}$
so, $\displaystyle f(x) = 5\cosh x + 3 \sinh x$
$\displaystyle f'(x) = 5\sinh x + 3\cosh x = \frac{8e^x - 2e^{-x}}{2} = 4e^x - e^{-x} = \frac{4e^{2x} - 1}{e^x}$
we want to find $\displaystyle x$ such that $\displaystyle f'(x) = 0$
since $\displaystyle e^x$ is never zero, we will solve $\displaystyle 4e^{2x} - 1 = 0$.. solving, we shall obtain
$\displaystyle x = -\ln 2$... substitute that $\displaystyle x$ to find the minimum value..
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to check, $\displaystyle f''(x) = f(x) = 5\cosh x + 3 \sinh x = 4e^x + e^{-x} > 0 $ for any $\displaystyle x$.
thus, $\displaystyle x = -\ln 2$ gives the minimum value for $\displaystyle f(x)$